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 one year ago
I was asked to write a program that finds the solutions, if any, to a given quadratic equation. My code is below. I have two issues: First When i input for example, a=3, b=4, c=5 the result prints" The roots are nan and nan" Secondly, I am to ensure that division by 0 is not permitted and the discriminant b^24ac cannot be negative. I am having a hard time with this bit. Could you help me solve this part?
 one year ago
I was asked to write a program that finds the solutions, if any, to a given quadratic equation. My code is below. I have two issues: First When i input for example, a=3, b=4, c=5 the result prints" The roots are nan and nan" Secondly, I am to ensure that division by 0 is not permitted and the discriminant b^24ac cannot be negative. I am having a hard time with this bit. Could you help me solve this part?

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tyteen4a03
 one year ago
Best ResponseYou've already chosen the best response.0Is there any code handy?

bunyonb
 one year ago
Best ResponseYou've already chosen the best response.0@tyteen4a03 /* This program takes in the values of the three coefficients a, b, and c as a screen input from the user. It then determines the roots of the quadratic equation using the formula ax^2 + bx + c = 0. The two roots are then outputted using the 'cout' command. */ #include <iostream> #include <cmath> using namespace std; int main(){ /* Declaring and initializing our variables a, b,c a and disc which is short for the discriminant, positive_root and negative_root*/ float a,b,c,disc,positive_root,negative_root; cout << "Enter the 3 coefficients a, b, c : " << endl; cin>>a>>b>>c; if(!a){ if(!b) cout << "Both a and b cannot be 0 in ax^2 + bx + c = 0" << "\n"; else { disc=c/b; cout << "The solution of the linear equation is : " << d << endl; } } else { disc=b*b4*a*c; if(d>0) positive_root=(b+sqrt(d))/(2*a); negative_root=(bsqrt(d))/(2*a); cout << "The first root = " << positive_root << endl; cout << "The second root = " << negative_root << endl; } getch(); return 0; }

bunyonb
 one year ago
Best ResponseYou've already chosen the best response.0This is a modified code of someone else's
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